Integrand size = 12, antiderivative size = 230 \[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=-\frac {\arcsin (a+b x)^2}{x}-\frac {2 b \arcsin (a+b x) \log \left (1-\frac {e^{i \arcsin (a+b x)}}{i a-\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}}+\frac {2 b \arcsin (a+b x) \log \left (1-\frac {e^{i \arcsin (a+b x)}}{i a+\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}}+\frac {2 i b \operatorname {PolyLog}\left (2,\frac {e^{i \arcsin (a+b x)}}{i a-\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}}-\frac {2 i b \operatorname {PolyLog}\left (2,\frac {e^{i \arcsin (a+b x)}}{i a+\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}} \]
-arcsin(b*x+a)^2/x-2*b*arcsin(b*x+a)*ln(1-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/ (I*a-(-a^2+1)^(1/2)))/(-a^2+1)^(1/2)+2*b*arcsin(b*x+a)*ln(1-(I*(b*x+a)+(1- (b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))/(-a^2+1)^(1/2)+2*I*b*polylog(2,(I* (b*x+a)+(1-(b*x+a)^2)^(1/2))/(I*a-(-a^2+1)^(1/2)))/(-a^2+1)^(1/2)-2*I*b*po lylog(2,(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))/(-a^2+1)^(1/ 2)
Time = 0.16 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.90 \[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\frac {-\sqrt {-1+a^2} \arcsin (a+b x)^2+2 i b x \arcsin (a+b x) \left (\log \left (\frac {a-\sqrt {-1+a^2}+i e^{i \arcsin (a+b x)}}{a-\sqrt {-1+a^2}}\right )-\log \left (\frac {a+\sqrt {-1+a^2}+i e^{i \arcsin (a+b x)}}{a+\sqrt {-1+a^2}}\right )\right )+2 b x \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (a+b x)}}{-a+\sqrt {-1+a^2}}\right )-2 b x \operatorname {PolyLog}\left (2,-\frac {i e^{i \arcsin (a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2} x} \]
(-(Sqrt[-1 + a^2]*ArcSin[a + b*x]^2) + (2*I)*b*x*ArcSin[a + b*x]*(Log[(a - Sqrt[-1 + a^2] + I*E^(I*ArcSin[a + b*x]))/(a - Sqrt[-1 + a^2])] - Log[(a + Sqrt[-1 + a^2] + I*E^(I*ArcSin[a + b*x]))/(a + Sqrt[-1 + a^2])]) + 2*b*x *PolyLog[2, (I*E^(I*ArcSin[a + b*x]))/(-a + Sqrt[-1 + a^2])] - 2*b*x*PolyL og[2, ((-I)*E^(I*ArcSin[a + b*x]))/(a + Sqrt[-1 + a^2])])/(Sqrt[-1 + a^2]* x)
Time = 0.80 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.90, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5304, 27, 5242, 5272, 3042, 3804, 25, 2694, 27, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\arcsin (a+b x)^2}{x^2} \, dx\) |
| \(\Big \downarrow \) 5304 |
| \(\displaystyle \frac {\int \frac {\arcsin (a+b x)^2}{x^2}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b \int \frac {\arcsin (a+b x)^2}{b^2 x^2}d(a+b x)\) |
| \(\Big \downarrow \) 5242 |
| \(\displaystyle b \left (-2 \int -\frac {\arcsin (a+b x)}{b x \sqrt {1-(a+b x)^2}}d(a+b x)-\frac {\arcsin (a+b x)^2}{b x}\right )\) |
| \(\Big \downarrow \) 5272 |
| \(\displaystyle b \left (-2 \int -\frac {\arcsin (a+b x)}{b x}d\arcsin (a+b x)-\frac {\arcsin (a+b x)^2}{b x}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (-2 \int \frac {\arcsin (a+b x)}{a-\sin (\arcsin (a+b x))}d\arcsin (a+b x)-\frac {\arcsin (a+b x)^2}{b x}\right )\) |
| \(\Big \downarrow \) 3804 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}-4 \int -\frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{-2 e^{i \arcsin (a+b x)} a-i e^{2 i \arcsin (a+b x)}+i}d\arcsin (a+b x)\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \int \frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{-2 e^{i \arcsin (a+b x)} a-i e^{2 i \arcsin (a+b x)}+i}d\arcsin (a+b x)\right )\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \left (\frac {i \int -\frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{2 \left (a+i e^{i \arcsin (a+b x)}-\sqrt {a^2-1}\right )}d\arcsin (a+b x)}{\sqrt {a^2-1}}-\frac {i \int -\frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{2 \left (a+i e^{i \arcsin (a+b x)}+\sqrt {a^2-1}\right )}d\arcsin (a+b x)}{\sqrt {a^2-1}}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \left (\frac {i \int \frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{a+i e^{i \arcsin (a+b x)}+\sqrt {a^2-1}}d\arcsin (a+b x)}{2 \sqrt {a^2-1}}-\frac {i \int \frac {e^{i \arcsin (a+b x)} \arcsin (a+b x)}{a+i e^{i \arcsin (a+b x)}-\sqrt {a^2-1}}d\arcsin (a+b x)}{2 \sqrt {a^2-1}}\right )\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \left (\frac {i \left (\int \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a+\sqrt {a^2-1}}\right )d\arcsin (a+b x)-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{\sqrt {a^2-1}+a}\right )\right )}{2 \sqrt {a^2-1}}-\frac {i \left (\int \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )d\arcsin (a+b x)-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )\right )}{2 \sqrt {a^2-1}}\right )\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \left (\frac {i \left (-i \int e^{-i \arcsin (a+b x)} \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a+\sqrt {a^2-1}}\right )de^{i \arcsin (a+b x)}-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{\sqrt {a^2-1}+a}\right )\right )}{2 \sqrt {a^2-1}}-\frac {i \left (-i \int e^{-i \arcsin (a+b x)} \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )de^{i \arcsin (a+b x)}-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )\right )}{2 \sqrt {a^2-1}}\right )\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle b \left (-\frac {\arcsin (a+b x)^2}{b x}+4 \left (\frac {i \left (i \operatorname {PolyLog}\left (2,-\frac {i e^{i \arcsin (a+b x)}}{a+\sqrt {a^2-1}}\right )-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{\sqrt {a^2-1}+a}\right )\right )}{2 \sqrt {a^2-1}}-\frac {i \left (i \operatorname {PolyLog}\left (2,-\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )-\arcsin (a+b x) \log \left (1+\frac {i e^{i \arcsin (a+b x)}}{a-\sqrt {a^2-1}}\right )\right )}{2 \sqrt {a^2-1}}\right )\right )\) |
b*(-(ArcSin[a + b*x]^2/(b*x)) + 4*(((-1/2*I)*(-(ArcSin[a + b*x]*Log[1 + (I *E^(I*ArcSin[a + b*x]))/(a - Sqrt[-1 + a^2])]) + I*PolyLog[2, ((-I)*E^(I*A rcSin[a + b*x]))/(a - Sqrt[-1 + a^2])]))/Sqrt[-1 + a^2] + ((I/2)*(-(ArcSin [a + b*x]*Log[1 + (I*E^(I*ArcSin[a + b*x]))/(a + Sqrt[-1 + a^2])]) + I*Pol yLog[2, ((-I)*E^(I*ArcSin[a + b*x]))/(a + Sqrt[-1 + a^2])]))/Sqrt[-1 + a^2 ]))
3.2.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) *(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q) Int[(f + g*x) ^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy mbol] :> Simp[2 Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x )) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ [a^2 - b^2, 0] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 1))), x] - Simp[b*c*(n/(e*(m + 1))) Int[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sq rt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[1/(c^(m + 1)*Sqrt[d]) Subst[In t[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c , d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (G tQ[m, 0] || IGtQ[n, 0])
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.91 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(b \left (-\frac {\arcsin \left (b x +a \right )^{2}}{b x}-\frac {2 \arcsin \left (b x +a \right ) \sqrt {-a^{2}+1}\, \left (\ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )-\ln \left (\frac {-i a +\sqrt {-a^{2}+1}+i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}}{-i a +\sqrt {-a^{2}+1}}\right )\right )}{a^{2}-1}+\frac {2 i \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {2 i \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {-i a +\sqrt {-a^{2}+1}+i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}}{-i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}\right )\) | \(301\) |
| default | \(b \left (-\frac {\arcsin \left (b x +a \right )^{2}}{b x}-\frac {2 \arcsin \left (b x +a \right ) \sqrt {-a^{2}+1}\, \left (\ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )-\ln \left (\frac {-i a +\sqrt {-a^{2}+1}+i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}}{-i a +\sqrt {-a^{2}+1}}\right )\right )}{a^{2}-1}+\frac {2 i \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {2 i \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {-i a +\sqrt {-a^{2}+1}+i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}}{-i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}\right )\) | \(301\) |
b*(-arcsin(b*x+a)^2/b/x-2*arcsin(b*x+a)*(-a^2+1)^(1/2)*(ln((I*a+(-a^2+1)^( 1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))-ln((-I*a+(-a^2+1 )^(1/2)+I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(-I*a+(-a^2+1)^(1/2))))/(a^2-1)+2*I *(-a^2+1)^(1/2)/(a^2-1)*dilog((I*a+(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^ (1/2))/(I*a+(-a^2+1)^(1/2)))-2*I*(-a^2+1)^(1/2)/(a^2-1)*dilog((-I*a+(-a^2+ 1)^(1/2)+I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(-I*a+(-a^2+1)^(1/2))))
\[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\int { \frac {\arcsin \left (b x + a\right )^{2}}{x^{2}} \,d x } \]
\[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\int \frac {\operatorname {asin}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \]
Exception generated. \[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for more details)Is
\[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\int { \frac {\arcsin \left (b x + a\right )^{2}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\arcsin (a+b x)^2}{x^2} \, dx=\int \frac {{\mathrm {asin}\left (a+b\,x\right )}^2}{x^2} \,d x \]